1) A median of a triangle is a segment from a vertex to the midpoint of the opposite side. The medians of any triangle are concurrent at the centroid. 2) An altitude of a triangle is a perpendicular segment from a vertex to the opposite side. The three altitudes of a triangle are concurrent at the orthocenter. 3) This document discusses properties of medians and altitudes of triangles, including using the centroid and orthocenter to solve problems involving lengths of segments and finding coordinates of special points like the centroid and orthocenter. Examples are provided to illustrate these concepts.

1. UNIT 5.4 MEDIANSUNIT 5.4 MEDIANS
AND ALTITUDESAND ALTITUDES

2. Warm Up
1. What is the name of the point where the angle
bisectors of a triangle intersect?
Find the midpoint of the segment with the
given endpoints.
2. (–1, 6) and (3, 0)
3. (–7, 2) and (–3, –8)
4. Write an equation of the line containing the
points (3, 1) and (2, 10) in point-slope form.
incenter
(–5, –3)
y – 1 = –9(x – 3)
(1, 3)

3. Apply properties of medians of a
triangle.
Apply properties of altitudes of a
triangle.
Objectives

4. median of a triangle
centroid of a triangle
altitude of a triangle
orthocenter of a triangle
Vocabulary

5. A median of a triangle is a segment whose
endpoints are a vertex of the triangle and the
midpoint of the opposite side.
Every triangle has three medians, and the medians
are concurrent.

6. The point of concurrency of the medians of a triangle
is the centroid of the triangle . The centroid is
always inside the triangle. The centroid is also called
the center of gravity because it is the point where a
triangular region will balance.

7. Example 1A: Using the Centroid to Find Segment
Lengths
In ∆LMN, RL = 21 and SQ =4.
Find LS.
LS = 14
Centroid Thm.
Substitute 21 for RL.
Simplify.

8. Example 1B: Using the Centroid to Find Segment
Lengths
In ∆LMN, RL = 21 and SQ =4.
Find NQ.
Centroid Thm.
NS + SQ = NQ Seg. Add. Post.
12 = NQ
Substitute 4 for SQ.
Multiply both sides by 3.
Substitute NQ for NS.
Subtract from both sides.

9. Check It Out! Example 1a
In ∆JKL, ZW = 7, and LX = 8.1.
Find KW.
Centroid Thm.
Substitute 7 for ZW.
Multiply both sides by 3.KW = 21

10. Check It Out! Example 1b
In ∆JKL, ZW = 7, and LX = 8.1.
Find LZ.
Centroid Thm.
Substitute 8.1 for LX.
Simplify.LZ = 5.4

11. Example 2: Problem-Solving Application
A sculptor is shaping a
triangular piece of iron that
will balance on the point of a
cone. At what coordinates will
the triangular region balance?

12. Example 2 Continued
11 Understand the Problem
The answer will be the coordinates of the
centroid of the triangle. The important
information is the location of the vertices,
A(6, 6), B(10, 7), and C(8, 2).
22 Make a Plan
The centroid of the triangle is the point of
intersection of the three medians. So write the
equations for two medians and find their point of
intersection.

13. Solve33
Let M be the midpoint of AB and N be the
midpoint of AC.
CM is vertical. Its equation is x = 8. BN has slope 1.
Its equation is y = x – 3. The coordinates of the
centroid are D(8, 5).
Example 2 Continued

14. Look Back44
Let L be the midpoint of BC. The equation for AL
is , which intersects x = 8 at D(8, 5).
Example 2 Continued

15. Check It Out! Example 2
Find the average of the x-coordinates and the
average of the y-coordinates of the vertices of
∆PQR. Make a conjecture about the centroid of a
triangle.

16. Check It Out! Example 2 Continued
The x-coordinates are 0, 6 and 3. The average
is 3.
The y-coordinates are 8, 4 and 0. The average
is 4.
The x-coordinate of the centroid is the average
of the x-coordinates of the vertices of the ∆,
and the y-coordinate of the centroid is the
average of the y-coordinates of the vertices of
the ∆.

17. An altitude of a triangle is a perpendicular segment
from a vertex to the line containing the opposite side.
Every triangle has three altitudes. An altitude can be
inside, outside, or on the triangle.

18. In ΔQRS, altitude QY is inside the triangle, but RX and
SZ are not. Notice that the lines containing the
altitudes are concurrent at P. This point of
concurrency is the orthocenter of the triangle.

19. The height of a triangle is the length of an
altitude.
Helpful Hint

20. Example 3: Finding the Orthocenter
Find the orthocenter of ∆XYZ with vertices
X(3, –2), Y(3, 6), and Z(7, 1).
Step 1 Graph the triangle.
X

21. Example 3 Continued
Step 2 Find an equation of the line containing the
altitude from Z to XY.
Since XY is vertical, the altitude is horizontal. The
line containing it must pass through Z(7, 1) so the
equation of the line is y = 1.

22. Example 3 Continued
Step 3 Find an equation of the line containing the
altitude from Y to XZ.
The slope of a line perpendicular to XZ is . This
line must pass through Y(3, 6).
Point-slope form.
Add 6 to both sides.
Substitute 6 for y1, for m,
and 3 for x1.
Distribute .

23. Step 4 Solve the system to find the coordinates of
the orthocenter.
The coordinates of the orthocenter are (6.75, 1).
Example 3 Continued
6.75 = x
Substitute 1 for y.
Subtract 10 from both sides.
Multiply both sides by

24. Check It Out! Example 3
Show that the altitude to JK passes through
the orthocenter of ∆JKL.
An equation of the altitude to JK is
Therefore, this altitude passes through the
orthocenter.
4 = 1 + 3
4 = 4 

25. Lesson Quiz
Use the figure for Items 1–3. In ∆ABC, AE = 12,
DG = 7, and BG = 9. Find each length.
1. AG
2. GC
3. GF
8
14
13.5
For Items 4 and 5, use ∆MNP with vertices M
(–4, –2), N (6, –2) , and P (–2, 10). Find the
coordinates of each point.
4. the centroid
5. the orthocenter
(0, 2)

26. All rights belong to their
respective owners.
Copyright Disclaimer Under
Section 107 of the
Copyright Act 1976,
allowance is made for "fair
use" for purposes such as
criticism, comment, news
reporting, TEACHING,
scholarship, and research.
Fair use is a use permitted
by copyright statute that
might otherwise be
infringing.
Non-profit, EDUCATIONAL
or personal use tips the
balance in favor of fair use.